3.2757 \(\int (c x)^m (a+b x^2)^2 \, dx\)

Optimal. Leaf size=58 \[ \frac {a^2 (c x)^{m+1}}{c (m+1)}+\frac {2 a b (c x)^{m+3}}{c^3 (m+3)}+\frac {b^2 (c x)^{m+5}}{c^5 (m+5)} \]

[Out]

a^2*(c*x)^(1+m)/c/(1+m)+2*a*b*(c*x)^(3+m)/c^3/(3+m)+b^2*(c*x)^(5+m)/c^5/(5+m)

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Rubi [A]  time = 0.03, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {270} \[ \frac {a^2 (c x)^{m+1}}{c (m+1)}+\frac {2 a b (c x)^{m+3}}{c^3 (m+3)}+\frac {b^2 (c x)^{m+5}}{c^5 (m+5)} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^m*(a + b*x^2)^2,x]

[Out]

(a^2*(c*x)^(1 + m))/(c*(1 + m)) + (2*a*b*(c*x)^(3 + m))/(c^3*(3 + m)) + (b^2*(c*x)^(5 + m))/(c^5*(5 + m))

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int (c x)^m \left (a+b x^2\right )^2 \, dx &=\int \left (a^2 (c x)^m+\frac {2 a b (c x)^{2+m}}{c^2}+\frac {b^2 (c x)^{4+m}}{c^4}\right ) \, dx\\ &=\frac {a^2 (c x)^{1+m}}{c (1+m)}+\frac {2 a b (c x)^{3+m}}{c^3 (3+m)}+\frac {b^2 (c x)^{5+m}}{c^5 (5+m)}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 41, normalized size = 0.71 \[ x (c x)^m \left (\frac {a^2}{m+1}+\frac {2 a b x^2}{m+3}+\frac {b^2 x^4}{m+5}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^m*(a + b*x^2)^2,x]

[Out]

x*(c*x)^m*(a^2/(1 + m) + (2*a*b*x^2)/(3 + m) + (b^2*x^4)/(5 + m))

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fricas [A]  time = 0.63, size = 87, normalized size = 1.50 \[ \frac {{\left ({\left (b^{2} m^{2} + 4 \, b^{2} m + 3 \, b^{2}\right )} x^{5} + 2 \, {\left (a b m^{2} + 6 \, a b m + 5 \, a b\right )} x^{3} + {\left (a^{2} m^{2} + 8 \, a^{2} m + 15 \, a^{2}\right )} x\right )} \left (c x\right )^{m}}{m^{3} + 9 \, m^{2} + 23 \, m + 15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^m*(b*x^2+a)^2,x, algorithm="fricas")

[Out]

((b^2*m^2 + 4*b^2*m + 3*b^2)*x^5 + 2*(a*b*m^2 + 6*a*b*m + 5*a*b)*x^3 + (a^2*m^2 + 8*a^2*m + 15*a^2)*x)*(c*x)^m
/(m^3 + 9*m^2 + 23*m + 15)

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giac [B]  time = 0.20, size = 135, normalized size = 2.33 \[ \frac {\left (c x\right )^{m} b^{2} m^{2} x^{5} + 4 \, \left (c x\right )^{m} b^{2} m x^{5} + 2 \, \left (c x\right )^{m} a b m^{2} x^{3} + 3 \, \left (c x\right )^{m} b^{2} x^{5} + 12 \, \left (c x\right )^{m} a b m x^{3} + \left (c x\right )^{m} a^{2} m^{2} x + 10 \, \left (c x\right )^{m} a b x^{3} + 8 \, \left (c x\right )^{m} a^{2} m x + 15 \, \left (c x\right )^{m} a^{2} x}{m^{3} + 9 \, m^{2} + 23 \, m + 15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^m*(b*x^2+a)^2,x, algorithm="giac")

[Out]

((c*x)^m*b^2*m^2*x^5 + 4*(c*x)^m*b^2*m*x^5 + 2*(c*x)^m*a*b*m^2*x^3 + 3*(c*x)^m*b^2*x^5 + 12*(c*x)^m*a*b*m*x^3
+ (c*x)^m*a^2*m^2*x + 10*(c*x)^m*a*b*x^3 + 8*(c*x)^m*a^2*m*x + 15*(c*x)^m*a^2*x)/(m^3 + 9*m^2 + 23*m + 15)

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maple [A]  time = 0.00, size = 94, normalized size = 1.62 \[ \frac {\left (b^{2} m^{2} x^{4}+4 b^{2} m \,x^{4}+2 a b \,m^{2} x^{2}+3 b^{2} x^{4}+12 a b m \,x^{2}+a^{2} m^{2}+10 a b \,x^{2}+8 a^{2} m +15 a^{2}\right ) x \left (c x \right )^{m}}{\left (m +5\right ) \left (m +3\right ) \left (m +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^m*(b*x^2+a)^2,x)

[Out]

x*(b^2*m^2*x^4+4*b^2*m*x^4+2*a*b*m^2*x^2+3*b^2*x^4+12*a*b*m*x^2+a^2*m^2+10*a*b*x^2+8*a^2*m+15*a^2)*(c*x)^m/(m+
5)/(m+3)/(m+1)

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maxima [A]  time = 0.61, size = 56, normalized size = 0.97 \[ \frac {b^{2} c^{m} x^{5} x^{m}}{m + 5} + \frac {2 \, a b c^{m} x^{3} x^{m}}{m + 3} + \frac {\left (c x\right )^{m + 1} a^{2}}{c {\left (m + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^m*(b*x^2+a)^2,x, algorithm="maxima")

[Out]

b^2*c^m*x^5*x^m/(m + 5) + 2*a*b*c^m*x^3*x^m/(m + 3) + (c*x)^(m + 1)*a^2/(c*(m + 1))

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mupad [B]  time = 1.20, size = 95, normalized size = 1.64 \[ {\left (c\,x\right )}^m\,\left (\frac {a^2\,x\,\left (m^2+8\,m+15\right )}{m^3+9\,m^2+23\,m+15}+\frac {b^2\,x^5\,\left (m^2+4\,m+3\right )}{m^3+9\,m^2+23\,m+15}+\frac {2\,a\,b\,x^3\,\left (m^2+6\,m+5\right )}{m^3+9\,m^2+23\,m+15}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^m*(a + b*x^2)^2,x)

[Out]

(c*x)^m*((a^2*x*(8*m + m^2 + 15))/(23*m + 9*m^2 + m^3 + 15) + (b^2*x^5*(4*m + m^2 + 3))/(23*m + 9*m^2 + m^3 +
15) + (2*a*b*x^3*(6*m + m^2 + 5))/(23*m + 9*m^2 + m^3 + 15))

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sympy [A]  time = 0.90, size = 345, normalized size = 5.95 \[ \begin {cases} \frac {- \frac {a^{2}}{4 x^{4}} - \frac {a b}{x^{2}} + b^{2} \log {\relax (x )}}{c^{5}} & \text {for}\: m = -5 \\\frac {- \frac {a^{2}}{2 x^{2}} + 2 a b \log {\relax (x )} + \frac {b^{2} x^{2}}{2}}{c^{3}} & \text {for}\: m = -3 \\\frac {a^{2} \log {\relax (x )} + a b x^{2} + \frac {b^{2} x^{4}}{4}}{c} & \text {for}\: m = -1 \\\frac {a^{2} c^{m} m^{2} x x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac {8 a^{2} c^{m} m x x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac {15 a^{2} c^{m} x x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac {2 a b c^{m} m^{2} x^{3} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac {12 a b c^{m} m x^{3} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac {10 a b c^{m} x^{3} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac {b^{2} c^{m} m^{2} x^{5} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac {4 b^{2} c^{m} m x^{5} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} + \frac {3 b^{2} c^{m} x^{5} x^{m}}{m^{3} + 9 m^{2} + 23 m + 15} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**m*(b*x**2+a)**2,x)

[Out]

Piecewise(((-a**2/(4*x**4) - a*b/x**2 + b**2*log(x))/c**5, Eq(m, -5)), ((-a**2/(2*x**2) + 2*a*b*log(x) + b**2*
x**2/2)/c**3, Eq(m, -3)), ((a**2*log(x) + a*b*x**2 + b**2*x**4/4)/c, Eq(m, -1)), (a**2*c**m*m**2*x*x**m/(m**3
+ 9*m**2 + 23*m + 15) + 8*a**2*c**m*m*x*x**m/(m**3 + 9*m**2 + 23*m + 15) + 15*a**2*c**m*x*x**m/(m**3 + 9*m**2
+ 23*m + 15) + 2*a*b*c**m*m**2*x**3*x**m/(m**3 + 9*m**2 + 23*m + 15) + 12*a*b*c**m*m*x**3*x**m/(m**3 + 9*m**2
+ 23*m + 15) + 10*a*b*c**m*x**3*x**m/(m**3 + 9*m**2 + 23*m + 15) + b**2*c**m*m**2*x**5*x**m/(m**3 + 9*m**2 + 2
3*m + 15) + 4*b**2*c**m*m*x**5*x**m/(m**3 + 9*m**2 + 23*m + 15) + 3*b**2*c**m*x**5*x**m/(m**3 + 9*m**2 + 23*m
+ 15), True))

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